# I have a problem from Karush-Kuhn-Tucker condition

maximize: $$xy-x$$ and s.t. $$x^2+y^2 \leq 9$$ and $$x \ge 0$$

Then, I tried to find points using lagrange method. I used 4 cases and they are $$(1)$$ $$\lambda_1 =0$$ and $$\lambda_2 = 0$$

forrect?

Maximize $$xy - x$$, given constraints: $$x^2 + y^2 \leq 9, x \ge 0$$

Taking $$xy - x = \lambda (x^2 + y^2 -9)$$

Taking derivative wrt $$x$$ and $$y$$, we get -

$$y-1 = 2 \lambda x$$ ...(i)

$$x = 2 \lambda y$$ ...(ii)

From (ii), $$\lambda = \frac{x}{2y}$$ and substituting in (i)

we get $$y(y-1) = x^2 \implies y(y-1) + y^2 \leq 9$$

That gives, $$y^2 - \frac{y}{2} \leq \frac{9}{2} \implies (y - \frac{1}{4})^2 \leq \frac{73}{16}$$

So $$y \leq \frac{\sqrt 73 + 1}{4} \approx 2.386001\,$$ (only taking positive values as $$x \ge 0$$ and we need $$xy$$ to be positive for max value)

taking max value of $$y$$, gives $$x \approx 1.818516$$

And max value of $$xy-x \approx 2.520465$$

• @Jumbo09 I did not use $\lambda_1$ and $\lambda_2$ and both constraints as it would complicate it. $x \geq 0$ is a check I did when deciding which values to consider$. – Math Lover Oct 21 '20 at 8:56 • @Jumbo09 In other words if the solution gave us certain values where$x$was negative, I would ignore those negative one's instead of adding it to the constraint upfront. – Math Lover Oct 21 '20 at 9:07 • ,so Which points are failed ? I didn't get it – Jumbo09 Oct 21 '20 at 9:19 • I understood this part , but I mean that when I solve this optimization problem, Points which are failed on constraint qualification become the candidate. Therefore, for this fail, what should I write the candidates in order to find the maximum. – Jumbo09 Oct 21 '20 at 9:35 • @Jumbo09 ok I understand you now. Your case 2,$\lambda_2 = 0, \lambda_1 \geq 0\$ should have given your maximum. Please check where you did a mistake in calculation. That case is very similar to what my solution is. – Math Lover Oct 21 '20 at 9:42